# Determine the equation of the the circle whose center is (3,-2) and the diameter is 8.

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We have the center (3, -2) and diameter = 8

We know that the formula for the circle is:

(y-y1)^2 + (x-x1)^2 = r^2

where (x1,y1) is the canter and r is the radius:

we hav teh diameter = 8

==> the radius r= diameter/2 = 8/2 = 4

==> r^2 = 16

**==> (y+2)^2 + (x-3)^2 = 16**

==> y^2 + 4y + 4 + x^2 - 6x + 9 = 16

==> y^2 + x^2 + 4y - 6x + 13 = 16

**==> y^2 + x^2 + 4y - 6x = 3**

We know that if the centre of the circle is (h,k) and the radius is r, then the equation of the circle is (x-h)^2 +(y-k)^2 = r^2.

Now we are given that the center is (3,-2) and diameter is 8. Therefore the radius is diameter /2 = 8/2 = 4.

Therefore the equationof the circle is got by substituting (3, -1) in place of (h,k) and 4 in place of r in the formula (x-h)^2+(y-k)^2 = r^2:

(x-3)^2 +(y-(-2))^2 = 4^2.

(x-3)^2 +(y+2)^2 = 16.

x^2-6x+9 +y^2 +4y +4 = 16 = 0.

x^2+y^2 -6x+4y+ 9+4-16 = 0.

x^2+y^2-6x+4y - 3 = 0 is the equation of the circle whose centre is (3, -2) and diameter is 8.